Chapter 8, Problem 8.13QA

Interpretation Introduction

To find:

The molarity of ${Na}^{+}$ ions in each of the following solutions:

a) $0.29\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{N}{\mathrm{O}}_3$

b) $0.33\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }25\mathrm{ }\mathrm{m}\mathrm{L}\mathrm{ }$ of solution

c) $0.88\mathrm{ }\mathrm{M}\mathrm{ }{\mathrm{N}\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$

d) $0.46\mathrm{ }\mathrm{g}{\mathrm{ }\mathrm{N}\mathrm{a}}_3\mathrm{P}{\mathrm{O}}_4\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }100\mathrm{ }\mathrm{m}\mathrm{L}$ of solution

Answer to Problem 8.13QA

Solution:

a) The molarity of ${Na}^{+}$ ions in $0.29 M NaN{O}_3$ is $0.29 M$.

b) The molarity of ${Na}^{+}$ ions in $0.33 g NaCl in 25 mL$ of solution is $0.23 M$.

c) The molarity of ${Na}^{+}$ ions in $0.88 M {Na}_{2}S{O}_4$ is $1.76 M$.

d) The molarity of ${Na}^{+}$ ions in $0.46 g{ Na}_{3}P{O}_4 in 100 mL$ of solution is $0.084 M$.

Explanation of Solution

1) Concept

To determine the concentration of ions present in the molecules, we need to use the moles of the compound and the mole ratio between them. We can calculate the moles from molarity and volume.

2) Formula

i) $Molarity= \frac{Moles of solute}{Volume in L}$

ii) $Molar mass=\frac{mass}{L of solution}$

3) Given

i) $0.29\mathrm{ }\mathrm{M}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{N}{\mathrm{O}}_3$

ii) $0.33\mathrm{ }\mathrm{g}\mathrm{ }\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }25\mathrm{ }\mathrm{m}\mathrm{L}\mathrm{ }$ of solution

iii) $0.88\mathrm{ }\mathrm{M}\mathrm{ }{\mathrm{N}\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$

iv) $0.46\mathrm{ }\mathrm{g}{\mathrm{ }\mathrm{N}\mathrm{a}}_3\mathrm{P}{\mathrm{O}}_4\mathrm{ }\mathrm{i}\mathrm{n}\mathrm{ }100\mathrm{ }\mathrm{m}\mathrm{L}$ of solution

4) Calculation

The molarity of ${Na}^{+}$ ions in each of the following solutions:

a) $0.29 M NaN{O}_3$

Dissociation equation for $NaN{O}_3$ is

$NaN{O}_3\to {Na}^{+}\left(aq\right)+N{O}_3^{-}\left(aq\right)$

The mole ratio of $1 mol NaN{O}_3: 1 mol {Na}^{+}$. We can use the volume of the solution as 1 L.

So  $0.29 mol NaN{O}_3=1 L solution$

$0.29 mol NaN{O}_3 \times \frac{1 mol {Na}^{+}}{1 mol NaN{O}_3}=0.29 mol {Na}^{+}$

So, $0.29 mol {Na}^{+}=1 L solution$

$Molarity of {Na}^{+}= \frac{0.29 mol {Na}^{+}}{1 L solution} =0.29 M$

Molarity of ${Na}^{+}$ ion is $0.29 M$.

b) $0.33 g NaCl in 25 mL of solution$

$0.33 g NaCl\times \frac{1 mol }{58.443 g}=0.00565 mol NaCl$

Dissociation of $NaCl$ is

$NaCl\to {Na}^{+}\left(aq\right)+{Cl}^{-} \left(aq\right)$

The mole ratio of $1 mol NaCl: 1 mol {Na}^{+}$

$0.00565 mol NaCl \times \frac{1 mol {Na}^{+}}{1 mol NaCl}=0.00565 mol {Na}^{+}$

$Volume= 25 mL\times \frac{1 L}{1000 ml}=0.025 L$

$Molarity of {Na}^{+}= \frac{0.00565 mol}{0.025 L} =0.2259 M {Na}^{+}$

$Molarity of {Na}^{+}=0.23 M$

Therefore, molarity of ${Na}^{+}$ ion is $0.23 M$.

c) $0.88 M {Na}_{2}S{O}_4$

Dissociation of ${Na}_{2}S{O}_4$ is

${Na}_{2}S{O}_4\to 2{Na}^{+}+{SO}_4^{2-}$

Since the mole ratio of ${Na}_{2}S{O}_4:{Na}^{+}$ is 1:2, we can use the volume of the solution as 1 L.

So, $0.88 mol {Na}_{2}S{O}_4=1 L solution$

$0.88 mol{Na}_{2}S{O}_4\times \frac{2 mol {Na}^{+}}{{1 mol Na}_{2}S{O}_4}=1.76 mol {Na}^{+}$

So, $1.76 mol {Na}^{+}$= 1 L of solution.

$Molarity of {Na}^{+} ion = \frac{1.76 mol {Na}^{+}}{1 L of solution}=1.76 M {Na}^{+}$

d) $0.46 g{ Na}_{3}P{O}_4 in 100 mL$ of solution

$0.46 g{ Na}_{3}P{O}_4\times \frac{1 mol}{163.94 g }=0.00281 mol { Na}_{3}P{O}_4$

Dissociation equation for ${ Na}_{3}P{O}_4$ is

${ Na}_{3}P{O}_4\to 3 {Na}^{+}+{PO}_4^{3-}$

The mole ratio of ${ Na}_{3}P{O}_4:{Na}^{+}$ is 1:3.

$0.00281 mol{ Na}_{3}P{O}_4\times \frac{3 mol {Na}^{+}}{1 mol { Na}_{3}P{O}_4}=0.00843 mol {Na}^{+}$

$Volume= 100 mL\times \frac{1 L}{1000 ml}=0.1 L$

$Molarity of {Na}^{+} ion = \frac{0.00843 M {Na}^{+}}{0.1 L solution}=0.0843 M$

$Molarity of {Na}^{+} ion=0.084 M$

Conclusion:

We calculated the molarity of Na⁺ from the molarity and moles ratios present in the compound and also from the mass and volume of the compounds.